Question 11, chap 15, sect 4 part 1 of 2 10 points A 1.24 kg block at rest on a

Question

Question 11, chap 15, sect 4 part 1 of 2 10 points A 1.24 kg block at rest on a tabletop is at- tached to a horizontal spring having constant 19.2 N/m. The spring is initially unstretched. A constant 29.8 N horizontal force is applied to the object causing the spring to stretch. The acceleration of gravity is 9.8 m/s 19.2 N/m 9.8 N Find the speed of the block after it has moved 0.238 m from equilibrium if the surface between block and tabletop is frictionless. Answer in units of m/s. Question 12, chap 15, sect 4. part 2 of 2 10 points

Explanation / Answer

Given,

m = 1.24 kg ; k = 19.2 N/m ; F = 29.8 N ; d = 0.238 m

Work done by the force will be:

W = Fd = 29.8 x .238 = 7.09 J

Now from energy conservation

W = 1/2 m v^2 + 1/2 k x^2

1/2 m v^2 = 7.09 - 0.5 x 19.2 x 0.238^2 = 7.09 - 6.55 = 6.55

v = sqrt (2 x 6.55/m) = sqrt(2 x 6.55/1.5) = 2.96 m/s

Hence, v = 2.96 m/s

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